How many elements does $\mathbb Z_7[i]/\langle i+1\rangle$ have? 
How many elements have $\mathbb Z_7[i]/\langle i+1\rangle$ ?

Elements of $Z_7[i]$ are of the form $a+bi$
$i+1$ is considered as zero in the quotient;
$i+1=0\iff i=-1\iff -1=i^2=1$
does it not contain now at most $2$ elements ?
or do I have to divide $a+bi$ by $1+i$;
$a+bi=a(1+i)+b-a\equiv b-a \pmod{1+i}$
thus the elements have no imaginary part, and I can get all $\{0,1,2,3,4,5,6\}$ for appropriate combinations of $a$ and $b$ ?
 A: Assuming 
$$\Bbb Z_7[i]=\Bbb F_{7^2}=\Bbb F_7[x]/\langle\,x^2+1\,\rangle =\text{ the field with $\;49\;$ elements}$$
Since a field only has the two trivial ideals (in fact, this characterizes fields among integer domains), and since clearly $\;\langle\,1+i\,\rangle\;$ is not the zero ideal, it then must be the whole ring, and finally
$$\Bbb F_{49}/\langle\,1+i\,\rangle\cong\{0\}$$
Remark: The above is a completely new, re-written answer to which the comments below do not apply.
A: In a comment, you say that you took a question about $\Bbb F_5[i]$ and modified it. This is the core of our difficulty. By $\Bbb F_5[i]$ I’m sure the author meant $\Bbb F_5[X]/(X^2+1)$. But this ring is not a field. So the question makes very good sense: calling $i$ the image of $X$ in this ring, we see that $1+i$ generates a group of order $4$. Since $1+i$ is a unit in our ring, the ideal it generates is the whole ring.
The situation is entirely different when we replace “$5$” by “$7$”, for $\Bbb F_5$ already has a square root of $-1$, whereas $\Bbb F_7$ does not. So, when we take $\Bbb F_7[X]/(X^2+1)$, we are defining a field. Our textbook-author’s question is much less apposite in this situation.
A: You're almost done: $\,\ {-}1\equiv 1\,\Rightarrow\,\color{#0a0}2\equiv 0\equiv \color{#0a0}7\,\Rightarrow\, \color{#c00}1 =  \color{#0a0}7-3\cdot \color{#0a0}2\equiv 0\,\Rightarrow\, R = \color{#c00}1\cdot R = \{0\}$
Remark $\ $ Generally in a ring of characteristic $\,\color{#0a0}m$ if we deduce that $\,\color{#0a0}{n=0}\,$ for $\,n\,$ coprime to $\,m\,$ then, by Bezout, $\, 1 = \gcd(m,n) = j \color{#0a0}m + k \color{#0a0}n  = 0,\,$ hence, as above, the ring is trivial.
